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Enigmatic patterns in Archimedean spirals



The Next CEO of Stack OverflowDivide circle into 9 pieces of equal areaLength and breadth of a rectangle enclosed between two semi-circles of given radiiIs there a plane algebraic curve with just 3-fold rotational symmetry, but without reflection symmetry?While calculating the arctan (1.01236) do we have to do 2 steps or one step before Taylor series?“Secondary spirals” within Fermat spiralsWhy is this family of dynamical systems able to produce spirals and clusters of points?How to prove the shape of the “Flower of Venus”Visualizing rational numbers as multiplication graphsThe slope of $nx % m$The quadrature of the circle: comparing Archimedean and Ulam spirals










4












$begingroup$


Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



The numbers are arranged by these formulas with $hat k = fracsqrtk2$



$$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
$$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



– the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



enter image description here
[Click image to enlarge.]



When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



enter image description here
[Click image to enlarge.]



When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



enter image description here
[Click image to enlarge.]



Here is my question:




How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
be explained? ("For $alpha=1$ there is a cross with a 8-fold
rotational symmetry, because ....")




Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



enter image description here




One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



enter image description here



enter image description here




To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



enter image description here



There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.










share|cite|improve this question











$endgroup$
















    4












    $begingroup$


    Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



    The numbers are arranged by these formulas with $hat k = fracsqrtk2$



    $$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
    $$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



    – the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



    This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



    enter image description here
    [Click image to enlarge.]



    When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



    enter image description here
    [Click image to enlarge.]



    When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



    enter image description here
    [Click image to enlarge.]



    Here is my question:




    How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
    be explained? ("For $alpha=1$ there is a cross with a 8-fold
    rotational symmetry, because ....")




    Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



    enter image description here




    One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



    In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



    enter image description here



    enter image description here




    To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



    enter image description here



    There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.










    share|cite|improve this question











    $endgroup$














      4












      4








      4


      3



      $begingroup$


      Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



      The numbers are arranged by these formulas with $hat k = fracsqrtk2$



      $$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
      $$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



      – the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



      This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



      enter image description here
      [Click image to enlarge.]



      When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



      enter image description here
      [Click image to enlarge.]



      When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



      enter image description here
      [Click image to enlarge.]



      Here is my question:




      How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
      be explained? ("For $alpha=1$ there is a cross with a 8-fold
      rotational symmetry, because ....")




      Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



      enter image description here




      One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



      In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



      enter image description here



      enter image description here




      To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



      enter image description here



      There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.










      share|cite|improve this question











      $endgroup$




      Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



      The numbers are arranged by these formulas with $hat k = fracsqrtk2$



      $$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
      $$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



      – the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



      This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



      enter image description here
      [Click image to enlarge.]



      When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



      enter image description here
      [Click image to enlarge.]



      When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



      enter image description here
      [Click image to enlarge.]



      Here is my question:




      How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
      be explained? ("For $alpha=1$ there is a cross with a 8-fold
      rotational symmetry, because ....")




      Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



      enter image description here




      One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



      In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



      enter image description here



      enter image description here




      To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



      enter image description here



      There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.







      geometry arithmetic visualization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago







      Hans-Peter Stricker

















      asked yesterday









      Hans-Peter StrickerHans-Peter Stricker

      6,74043995




      6,74043995




















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago


















          1












          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago


















          1












          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago












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          3 Answers
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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          2












          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago













          2












          2








          2





          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.







          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered yesterday









          MiltenMilten

          1335




          1335




          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.











          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago
















          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          $begingroup$
          Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago




          $begingroup$
          Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago












          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago




          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago











          1












          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          1












          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago













          1












          1








          1





          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$



          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 16 hours ago









          ChrystomathChrystomath

          1,858513




          1,858513











          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago
















          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago




          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago











          1












          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago
















          1












          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago














          1












          1








          1





          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$



          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 16 hours ago

























          answered 17 hours ago









          Hans-Peter StrickerHans-Peter Stricker

          6,74043995




          6,74043995











          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago

















          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago
















          $begingroup$
          Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
          $endgroup$
          – Milten
          17 hours ago





          $begingroup$
          Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
          $endgroup$
          – Milten
          17 hours ago













          $begingroup$
          I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
          $endgroup$
          – Milten
          17 hours ago




          $begingroup$
          I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
          $endgroup$
          – Milten
          17 hours ago












          $begingroup$
          Can you give an analytical expression for this requirement, please? I would add it to my answer.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago




          $begingroup$
          Can you give an analytical expression for this requirement, please? I would add it to my answer.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago












          $begingroup$
          Did it. But still there's the ugly gap between 23 and 9.
          $endgroup$
          – Hans-Peter Stricker
          16 hours ago




          $begingroup$
          Did it. But still there's the ugly gap between 23 and 9.
          $endgroup$
          – Hans-Peter Stricker
          16 hours ago












          $begingroup$
          Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
          $endgroup$
          – Milten
          16 hours ago





          $begingroup$
          Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
          $endgroup$
          – Milten
          16 hours ago


















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